A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150 meters of fencing for the other three sides. The area A(x) in square meters of the schoolyard is a function of the length x in meters of each of the sides perpendicular to the school wall. a. Write an expression for A(x).b. What is the area of the schoolyard when x=40?c. What is a reasonable domain for A in this context?

Answer:Part 1) The expression is [tex]A(x)=150x-2x^2[/tex]Part 2) The area of the schoolyard when x=40 m is A=2,800 m^2Part 3) The domain is all real numbers greater than zero and less than 75 metersStep-by-step explanation:Part 1) Write an expression for A(x)Letx -----> the length of the rectangular schoolyardy ---> the width of the rectangular schoolyardwe know thatThe perimeter of the fencing (using the wall of the school for one side) is[tex]P=2x+y[/tex][tex]P=150\ m[/tex]so[tex]150=2x+y[/tex][tex]y=150-2x[/tex] -----> equation AThe area of the rectangular schoolyard is[tex]A=xy[/tex] ---> equation Bsubstitute equation A in equation B[tex]A=x(150-2x)[/tex][tex]A=150x-2x^2[/tex]Convert to function notation[tex]A(x)=150x-2x^2[/tex]Part 2) What is the area of the schoolyard when x=40?For x=40 msubstitute in the expression of Part 1) and solve for A[tex]A(40)=150(40)-2(40^2)[/tex][tex]A(40)=2,800\ m^2)[/tex]Part 3) What is a reasonable domain for A(x) in this contextwe know thatA represent the area of the rectangular schoolyardx represent the length of of the rectangular schoolyardwe have[tex]A(x)=150x-2x^2[/tex]This is a vertical parabola open downwardThe vertex is a maximumThe x-coordinate of the vertex represent the length for the maximum areaThe y-coordinate of the vertex represent the maximum areaThe vertex is the point (37.5, 2812.5)using a graphing tool, see the attached figurethereforeThe maximum area is 2,812.5 m^2The x-intercepts are x=0 m and x=75 mThe domain for A is the interval -----> (0, 75)All real numbers greater than zero and less than 75 meters